91Ó°ÊÓ

• The spring constant is, $k=600N/m.$

• Mass of the putty is,$m_p=0.50\mathrm{kg}$ .
• The period of SHM executing by block is,$T_1=0.40\mathrm{s}$.
• The amplitude of SHM executing by block is, $x_m=0.20\mathrm{m}$.

The period T is the time required for one complete oscillation or cycle. It is related to the frequency by,

$\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

It is also related to mass (m) and force constant (k) by the formula,

localid="1657256993714" $\mathit{T}\mathbf{=}\mathbf{2}\mathbf{Ï€}\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

We can find the mass of the block from the period of its motion. Then using it and the mass of the putty in the formula for the period of SHM, we can find the new period of the motion of the block. Then using the law of conservation of momentum, we can find the maximum velocity of the putty-block system. Lastly, using the formula for the maximum velocity of SHM we can find the new amplitude of the motion of the block.

Formulae:

The period of oscillation for SHM,$T=2\mathrm{Ï€}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ (i)
Conservation of momentum gives,

Momentum before collision=momentum after collision (ii)

Using equation (i) and the value of the old period of oscillation, the mass of the block can be given as:

$$\begin{gathered} m_b=\frac{{T^2}_{1}k}{4{\mathrm{Ï€}}^2} \\ =\frac{{\left(0.40s\right)}^2(600\mathrm{N}/\mathrm{m})}{4{(3.142)}^2} \\ =2.43\mathrm{kg} \end{gathered}$$

Then, the total mass of the putty-block system is given by:

$$\begin{gathered} m_b+m_p=2.43\mathrm{kg}+0.50\mathrm{kg} \\ =2.93\mathrm{kg} \end{gathered}$$

Again, using equation (i) and the total mass of the system, the new period of oscillation is given as:

$$\begin{gathered} T_2=2(3.142)\sqrt{\frac{2.93\mathrm{kg}}{600\mathrm{N}/\mathrm{m}}} \\ =0.439\mathrm{s} \\ ≈0.44\mathrm{s} \end{gathered}$$

Therefore, the new period of the motion of the block is $0.44\mathrm{s}$.

The speed of the block before the collision is given as:

$$\begin{gathered} v_b=\mathrm{Ó\neg }{X}_m \\ =\frac{2\mathrm{Ï€}}{T^1}{X}_m(âˆ\mu \mathrm{angular}\mathrm{frequency},\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T_1}) \\ =\frac{2(3.142)(0.20\mathrm{m})}{0.40\mathrm{s}} \\ =3.142\mathrm{m}/\mathrm{s} \end{gathered}$$

The collision between block and putty is an inelastic collision.

According to equation (ii), we get the equation as:

$m_b{v}_b+m_p{v}_p=(m_b+m_p)v_f$

$(2.43\mathrm{kg})(3.142\mathrm{m}/\mathrm{s})+(0.50\mathrm{kg})\left(0\right)=(2.93)$

The amplitude of the motion of the putty-block system is given as:

$$\begin{gathered} X_{m.f}=\frac{v_f}{\mathrm{Ó\neg }} \\ =\frac{v_f}{2\mathrm{Ï€}/{\mathrm{T}}_2}(âˆ\mu \mathrm{angular}\mathrm{frequency},\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{{\mathrm{T}}_2} \\ =\frac{v_f{T}_2}{2\mathrm{Ï€}} \\ =\frac{2.61\mathrm{m}/\mathrm{s}(0.44\mathrm{s})}{\left(2\right)(3.142)} \\ =0.18\mathrm{m} \end{gathered}$$

Therefore, the new amplitude of the motion of block is 0.18m.